The question asks to simplify the following expression by writing it as a single summation (the second sigma in the expression should have upper limit of n+2):
$$3\sum_{i=2}^n (2i^2-i) – \sum_{j=4}^n (3j^2+1)$$
Im having trouble figuring this one out, I tried to write each inner term as a sum then use recursive definitions to expand them all then simplify:
$$ \sum_{i=2}^n (6i^2) – \sum_{i=2}^n (3i) – \sum_{j=4}^n (3j^2) + \sum_{j=4}^n 1$$
$$6(\frac{n(n+1)(2n+1)}{6}) – 3(\frac{n(n+1)}{2}) – 3(\frac{(n+2)((n+2)+1)(2(n+1)+1)}{6} + \frac{n(n+1)}{2}$$
from here I continued to simplify until it was in the simplest form but i dont understand how to express it a single summation, also I think I may have gone about this the wrong way and made some mistakes here and there. Is anyone able to correct me or point me in the right direction in terms of solving this?
-Thanks
Best Answer
(Assuming the upper limit is $n+2.$) You can change variables in a summation just like in an integral. Here, you want the limits to match, but the second sum's limits are $2$ larger than the first. So in the second sum set $i=j+2$ so that $j=i-2$. Then when $j=4$, $i=2$ and when $j=n+2$, $i=n$. The second sum becomes
$$\sum_{i=2}^{n} (3(i+2)^2 +1).$$
Now you can combine the sums and simplify the summand.
If the upper limit is not $n+2$, then you start by doing the same thing, but then there's some more work.